Match the following: Column I (A) In a...

MathematicsComponents of a VectorJEE Advanced 2015Difficult

Match the following:

List-I

(P)

In a triangle

Δ X Y Z

, let

a, b

, and

c

be the lengths of the sides opposite to the angles

X, Y

and

Z

, respectively. If

2 (a^{2} - b^{2}) = c^{2}

and

λ = \frac{s i n ( X - Y )}{s i n Z}

, then possible values of

n

for which

cos (nπ λ) = 0

is (are)

(Q)

In a triangle

Δ X Y Z

, let

a, b

and

c

be the lengths of the sides opposite to the angles

X, Y

, and

Z

respectively. If

1 + cos 2 X - 2 cos 2 Y = 2 sin X sin Y

, then possible value(s) of

\frac{a}{b}

is (are)

(R)

R^{2}

, let

3 \hat{i} + \hat{j}, \hat{i} + 3 \hat{j}

and

β \hat{i} + (1 - β) \hat{j}

be the position vectors of

X, Y

and

Z

with respect to the origin

O

, respectively. If the distance of

Z

from the bisector of the acute angle of

O X

with

O Y

\frac{3}{2}

, then possible value(s) of

∣ β ∣

is (are)

(S)

Suppose that

F (α)

denotes the area of the region bounded by

x = 0, x = 2, y^{2} = 4 x

and

y = ∣ α x - 1∣ + ∣ α x - 2∣ + α x

, where

α \in {0, 1}

. Then the value(s) of

F (α) + \frac{8}{3} 2

, when

α = 0

and

α = 1

, is (are)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → NaN, Q → 1, P → 2, S → 5

Visualized Solution (English)

Part A: Sine Rule Application

Given: $2 (a^{2} - b^{2}) = c^{2}$
Using Sine Rule: $a = 2 R tan X$ , $b = 2 R tan Y$ , $c = 2 R tan Z$
Substitute: $2 (4 R^{2} tan^{2} X - 4 R^{2} tan^{2} Y) = 4 R^{2} tan^{2} Z$
Simplify: $2 (tan^{2} X - tan^{2} Y) = tan^{2} Z$

Part A: Trigonometric Simplification

Identity: $sin^{2} A - sin^{2} B = sin (A - B) sin (A + B)$
Equation: $2 sin (X - Y) sin (X + Y) = sin^{2} Z$
Since $X + Y = π - Z$ , then $sin (X + Y) = sin Z$
Result: $2 sin (X - Y) sin Z = sin^{2} Z$

Part A: Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

$λ = \frac{s i n ( X - Y )}{s i n Z} = \frac{1}{2}$
Condition: $cos (nπ λ) = 0 ⟹ cos (\frac{nπ}{2}) = 0$
General Solution: $\frac{nπ}{2} = (2 k + 1) \frac{π}{2}$
Possible values of $n$ : $1, 3, 5, \dots$

Part B: Trigonometric Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a / b

Given: $1 + cos 2 X - 2 cos 2 Y = 2 sin X sin Y$
Substitute: $2 cos^{2} X - 2 (1 - 2 sin^{2} Y) = 2 sin X sin Y$
Simplify: $2 (1 - sin^{2} X) - 2 + 4 sin^{2} Y = 2 sin X sin Y$
Factor: $(2 sin Y + sin X) (sin Y - sin X) = 0$
Result: $sin Y = sin X ⟹ a = b ⟹ \frac{a}{b} = 1$

Part C: Vector Magnitudes and Bisector

$O X = 3 \hat{i} + \hat{j}$ , $O Y = \hat{i} + 3 \hat{j}$
Magnitudes: $∣ O X ∣ = ∣ O Y ∣ = 2$
Bisector direction: $O X + O Y = (3 + 1) (\hat{i} + \hat{j})$
Bisector Equation: $x - y = 0$

Part C: Distance Formula and Beta

Point $Z (β, 1 - β)$ , Line $x - y = 0$
Distance: $\frac{∣ β - ( 1 - β ) ∣}{1 ^{2} + ( - 1 ) ^{2}} = \frac{3}{2}$
Solve: $∣2 β - 1∣ = 3 ⟹ 2 β - 1 = \pm 3$
Values: $β = 2, - 1 ⟹ ∣ β ∣ = 2, 1$

Part D: Area for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α = 0

For $α = 0$ : $y = ∣ - 1∣ + ∣ - 2∣ + 0 = 3$
Area $F (0) = \int_{0}^{2} (3 - 2 x) d x$
Integration: $[3 x - \frac{4}{3} x^{3/2}]_{0}^{2} = 6 - \frac{8}{3} 2$
Result: $F (0) + \frac{8}{3} 2 = 6$

Part D: Area for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α = 1

For $α = 1$ : $y = ∣ x - 1∣ + ∣ x - 2∣ + x$
Piecewise: $y = 3 - x$ for $0 \leq x < 1$ and $y = x + 1$ for $1 \leq x \leq 2$
Area $F (1) = \int_{0}^{1} (3 - x - 2 x) d x + \int_{1}^{2} (x + 1 - 2 x) d x$
Result: $F (1) + \frac{8}{3} 2 = 5$

Final Conclusion and Matching

Final Matches:
(A) $\to 1, 3, 5$
(B) $\to 1$
(C) $\to 1, 2$
(D) $\to 5, 6$

00:00 / 00:00

Conceptually Similar Problems

MathematicsEquation of a PlaneJEE Advanced 2006Difficult

View in:English Hindi

Match the following :

List-I

(P)

Two rays

x + y = ∣ a ∣

and

a x - y = 1

intersects each other in the first quadrant in the interval

a \in (a_{0}, \infty)

, the value of

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Point

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. Let

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\hat{k} \times (\hat{k} \times a) = 0

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γ =

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sin A sin B sin C + cos A cos B = 1

, then the value of

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(1)

(2)

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(3)

\int_{0}^{1} 1 - x d x + \int_{- 1}^{0} 1 + x d x

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MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2010Difficult

View in:English Hindi

Match the statement in Column-I with the values in Column-II

List-I

(P)

A line from the origin meets the lines

\frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1}

and

\frac{x - \frac{8}{3}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1}

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Q

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, then

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The values of

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satisfying

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are

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Non-zero vectors

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and

c

satisfy

a \cdot b = 0

(b - a) \cdot (b + c) = 0

and

2∣ b + c ∣ = ∣ b - a ∣

. If

a = μ b + 4 c

, then the possible values of

μ

are

(S)

Let

f

be the function on

[- π, π]

given by

f (0) = 9

and

f (x) = \frac{s i n ( \frac{9 x}{2} )}{s i n ( \frac{x}{2} )}

for

x \neq = 0

. The value of

\frac{2}{π} \int_{- π}^{π} f (x) d x

List-II

(1)

-4

(2)

(3)

(4)

(5)

Answer:P → 5, P → 3, Q → 4, S → 3

MathematicsEquation of a Line in SpaceJEE Advanced 2006Moderate

View in:English Hindi

Match the following

List-I

(P)

\sum_{i = 1}^{\infty} tan^{- 1} (\frac{1}{2 i ^{2}}) = t

, then

tan t =

(Q)

Sides

a, b, c

of a triangle

A B C

are in AP and

cos θ_{1} = \frac{a}{b + c}

cos θ_{2} = \frac{b}{a + c}

cos θ_{3} = \frac{c}{a + b}

, then

tan^{2} (\frac{θ _{1}}{2}) + tan^{2} (\frac{θ _{3}}{2}) =

(R)

A line is perpendicular to

x + 2 y + 2 z = 0

and passes through

(0, 1, 0)

. The perpendicular distance of this line from the origin is

List-II

(1)

(2)

\frac{5}{3}

(3)

\frac{2}{3}

Answer:P → 1, Q → 3, R → 2

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View in:English Hindi

Match the following:

List-I

(P)

R^{2}

, if the magnitude of the projection vector of the vector

α \hat{i} + β \hat{j}

3 \hat{i} + \hat{j}

3

and if

α = 2 + 3 β

, then possible value of

∣ α ∣

is/are

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Let

a

and

b

be real numbers such that the function

f (x) = {- 3 a x^{2} - 2, b x + a^{2}, x < 1 x \geq 1

if differentiable for all

x \in R

. Then possible value of

a

is (are)

(R)

Let

ω \neq = 1

be a complex cube root of unity. If

(3 - 3 ω + 2 ω^{2})^{4 n + 3} + (2 + 3 ω - 3 ω^{2})^{4 n + 3} + (- 3 + 2 ω + 3 ω^{2})^{4 n + 3} = 0

, then possible value (s) of

n

is (are)

(S)

Let the harmonic mean of two positive real numbers

a

and

b

be 4. If

q

is a positive real number such that

a, 5, q, b

is an arithmetic progression, then the value(s) of

∣ q - a ∣

is (are)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → 2, P → 2, R → NaN, Q → 5

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View in:English Hindi

Match the statements given in Column-I with the values given in Column-II.

List-I

(P)

a = \hat{j} + 3 \hat{k}

b = - \hat{j} + 3 \hat{k}

and

c = 23 \hat{k}

form a triangle, then the internal angle of the triangle between

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and

b

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\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}

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(S)

The maximum value of

Arg (\frac{1}{1 - z})

for

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is given by

List-II

(1)

\frac{π}{6}

(2)

\frac{2 π}{3}

(3)

\frac{π}{3}

(4)

π

(5)

\frac{π}{2}

Answer:P → 2, Q → 1, R → 4, S → 5

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Match List I with List II:

List-I

(P)

Let

y (x) = cos (3 cos^{- 1} x), x \in [- 1, 1], x \neq = \pm \frac{3}{2}

. Then

\frac{1}{y ( x )} {(x^{2} - 1) \frac{d ^{2} y ( x )}{d x ^{2}} + x \frac{d y ( x )}{d x}}

equals

(Q)

Let

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be the vertices of a regular polygon of

n

sides with its centre at the origin. Let

a_{k}

be the position vector of the point

A_{k}, k = 1, 2, \dots, n

. If

∣ \sum_{k = 1}^{n - 1} (a_{k} \times a_{k + 1}) ∣ = ∣ \sum_{k = 1}^{n - 1} (a_{k} \cdot a_{k + 1}) ∣

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n

(R)

If the normal from the point

P (h, 1)

on the ellipse

\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

is perpendicular to the line

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, then the value of

h

(S)

Number of positive solutions satisfying the equation

tan^{- 1} (\frac{1}{2 x + 1}) + tan^{- 1} (\frac{1}{4 x + 1}) = tan^{- 1} (\frac{2}{x ^{2}})

List-II

(1)

(2)

(3)

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

MathematicsAlgebraic Operations on MatricesJEE Advanced 2008Moderate

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Match the Statements/Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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(P)

The minimum value of

\frac{x ^{2} + 2 x + 4}{x + 2}

(Q)

Let A and B be

3 \times 3

matrices of real numbers, where A is symmetric, B is skew-symmetric, and

(A + B) (A - B) = (A - B) (A + B)

. If

(A B)^{t} = (- 1)^{k} A B

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(A B)^{t}

is the transpose of the matrix AB, then the possible values of

k

are

(R)

Let

a = lo g_{3} lo g_{3} 2

. An integer

k

satisfying

1 < 2^{- k + 3^{- a}} < 2

, must be less than

(S)

sin θ = cos ϕ

, then the possible values of

\frac{1}{π} (θ \pm ϕ - \frac{π}{2})

are

List-II

(1)

(2)

(3)

(4)

Answer:P → 3, Q → 4, R → 4, P → 3

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 2011Moderate

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Comprehension Passage

Let

a, b

and

c

be three real numbers satisfying

[a b c] 187923777 = [000] \dots (E)

Question 1:

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

(A)

(B)

(C)

(D)

Answer:D

Question 2:

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

(A)

-2

(B)

(C)

(D)

-3

Answer:A

Question 3:

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

(A)

(B)

(C)

6/7

(D)

\infty

Answer:B

MathematicsScalar Triple ProductJEE Advanced 2009Moderate

View in:English Hindi

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

(P)

Root(s) of the equation

2 sin^{2} θ + sin^{2} 2 θ = 2

(Q)

Points of discontinuity of the function

f (x) = [\frac{6 x}{π}] cos [\frac{3 x}{π}]

, where

[y]

denotes the largest integer less than or equal to

y

(R)

Volume of the parallelopiped with its edges represented by the vectors

\hat{i} + \hat{j}, \hat{i} + 2 \hat{j}

and

\hat{i} + \hat{j} + π \hat{k}

(S)

Angle between vector

a

and

b

where

a, b

and

c

are unit vectors satisfying

a + b + 3 c = 0

List-II

(1)

π /6

(2)

π /4

(3)

π /3

(4)

π /2

(5)

π

Answer:Q → 4, Q → NaN, R → 5, S → 3

MathematicsSolving Inverse Trigonometric EquationsJEE Advanced 2013Moderate

View in:English Hindi

Match List I with List II:

List-I

(P)

[\frac{1}{y ^{2}} (\frac{c o s ( t a n ^{- 1} y ) + y s i n ( t a n ^{- 1} y )}{c o t ( s i n ^{- 1} y ) + t a n ( s i n ^{- 1} y )})^{2} + y^{4}]^{1/2}

takes value

(Q)

cos x + cos y + cos z = 0 = sin x + sin y + sin z

then possible value of

cos \frac{x - y}{2}

(R)

cos (\frac{π}{4} - x) cos 2 x + sin x sin 2 x sec x = cos x sin 2 x sec x + cos (\frac{π}{4} + x) cos 2 x

then possible value of

sec x

(S)

cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6))

x \neq = 0

, then possible value of

x

List-II

(1)

\frac{1}{2} \frac{5}{3}

(2)

2

(3)

\frac{1}{2}

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

Match the following:

List-I

List-II

Visualized Solution (English)

Part A: Sine Rule Application

Part A: Trigonometric Simplification

Part A: Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Part B: Trigonometric Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a/b

Part C: Vector Magnitudes and Bisector

Part C: Distance Formula and Beta

Part D: Area for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=0

Part D: Area for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=1

Final Conclusion and Matching

Conceptually Similar Problems

Match the following :

List-I

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Match the statement in Column-I with the values in Column-II

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Match the following

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Match the statements given in Column-I with the values given in Column-II.

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Match List I with List II:

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the point P(a,b,c), with reference to (E), lies on the plane 2x+y+z=1, then the value of 7a+b+c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ω be a solution of x3−1=0 with Im(ω)>0, if a=2 with b and c satisfying (E), then the value of ωa3​+ωb1​+ωc3​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let b=6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2+bx+c=0, then ∑n=0∞​(α1​+β1​)n is

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

List-II

Match List I with List II:

List-I

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Match the following:

List-I

List-II

Match the following :

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List-II

Match the statement in Column-I with the values in Column-II

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List-II

Match the following

List-I

List-II

Match the following:

List-I

List-II

Match the statements given in Column-I with the values given in Column-II.

List-I

List-II

Match List I with List II:

List-I

List-II

List-I

List-II

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the point P(a,b,c), with reference to (E), lies on the plane 2x+y+z=1, then the value of 7a+b+c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ω be a solution of x3−1=0 with Im(ω)>0, if a=2 with b and c satisfying (E), then the value of ωa3​+ωb1​+ωc3​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let b=6, with a and c satisfying (E). If α and β are the roots of the quadratic equation ax2+bx+c=0, then ∑n=0∞​(α1​+β1​)n is

Match the statements / expressions given in Column-I with the values given in Column-II.

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Part A: Solving for $λ$ and $n$

Part B: Trigonometric Ratio $a / b$

Part D: Area for $α = 0$

Part D: Area for $α = 1$

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is